检测哪个按钮已被点击并用ajx提交

Question

我需要帮助获取点击按钮的值并使用AJAX发送此信息。这两个脚本单独工作。但我需要将“buttonvalue”传递给AJAX“data”值“action”。我正在使用的是以下。

$("document").ready(function(){ 

    $(".js-ajax-php-json").on("click", "button[name=mysqljob]", function(){ 
    var buttonvalue = $(this).attr("value"); 
    alert(buttonvalue); 
    }); 

    $(".js-ajax-php-json").submit(function(){ 

    var data = {"action": buttonvalue}; 

    data = $(this).serialize() + "&" + $.param(data); 

    $.ajax({ 
     type: "POST", 
     dataType: "json", 
     url: "phpVars/ajaxUpload.php", //Relative or absolute path to response.php file 
     data: data, 
     success: function(data) { 
     $(".the-return").html(
     data["form"] 
     ); 
     // alert("Form submitted successfully.\nReturned json: " + data["json"]); 
     } 
    }); 
    return false; 
    }); 
}); 

Answer 1

我会阻止表单提交做你想要的然后通过ajax调用发布数据。请尝试以下操作：

$("document").ready(function(){ 

    $(".js-ajax-php-json").on("click", "button[name=mysqljob]", function (e){ 
     e.preventDefault(); 
     var buttonvalue = $(this).attr("value"); 
     var data = {"action": buttonvalue, "formData": $(this).serialize() + "&" + $.param(data) }; 

     $.ajax({ 
      type: "POST", 
      dataType: "json", 
      url: "phpVars/ajaxUpload.php", //Relative or absolute path to response.php file 
      data: data, 
      success: function(data) { 
       $(".the-return").html(
        data["form"] 
       ); 
    //   alert("Form submitted successfully.\nReturned json: " + data["json"]); 
      } 
     }); 
    return false; 
    }); 
}); 

未测试。